The electric field between two charged parallel platesseparated by a distance of

Question

The electric field between two charged parallel platesseparated by a distance of 9.0 mm has a uniform value of40,000 N/C. 1) What is the potential difference between the plates? 2) How much kinetic energy would an electron gain if itaccelerated from the negative plate to the positive plate? The electric field between two charged parallel platesseparated by a distance of 9.0 mm has a uniform value of40,000 N/C. 1) What is the potential difference between the plates? 2) How much kinetic energy would an electron gain if itaccelerated from the negative plate to the positive plate?

Explanation / Answer

The electricfield between two charged parallel plates separated by adistance
of 9.0 mm has a uniform value of 40,000N/C.

(1) What is the potential difference betweenthe plates?
    V = E*d = (40000N/C)*(9.0e-3 m) = 360 Volts

(2) How much kinetic energy would anelectron gain if it accelerated from the
negative plate to the positive plate?
    {K.E.} = Q*V = (1.602e-19 C)*(360 V) = 5.767e-17 Joules

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